Question: Simplify; express your answer in exponential form. Assume $n\neq 0, p\neq 0$. $\dfrac{{(n^{-4})^{-1}}}{{(np^{-3})^{-3}}}$
Answer: To start, try working on the numerator and the denominator independently. In the numerator, we have ${n^{-4}}$ to the exponent ${-1}$ . Now ${-4 \times -1 = 4}$ , so ${(n^{-4})^{-1} = n^{4}}$ In the denominator, we can use the distributive property of exponents. ${(np^{-3})^{-3} = (n)^{-3}(p^{-3})^{-3}}$ Simplify using the same method from the numerator and put the entire equation together. $\dfrac{{(n^{-4})^{-1}}}{{(np^{-3})^{-3}}} = \dfrac{{n^{4}}}{{n^{-3}p^{9}}}$ Break up the equation by variable and simplify. $\dfrac{{n^{4}}}{{n^{-3}p^{9}}} = \dfrac{{n^{4}}}{{n^{-3}}} \cdot \dfrac{{1}}{{p^{9}}} = n^{{4} - {(-3)}} \cdot p^{- {9}} = n^{7}p^{-9}$.